Leetcode#746. Min Cost Climbing Stairs(最低花费登楼梯--动态规划)

题目

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Example 1:

Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2:

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].

Note:

cost will have a length in the range [2, 1000].

Every cost[i] will be an integer in the range [0, 999].

题意

每一个楼梯上花费为cost[i]，你花费cost[i]可以登上1个楼梯或者1个楼梯，

同样，开始的时候你可以选择登上1或者登上2个楼梯。

题解

动态规划，划分成局部最优考虑。

C++代码

class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int dp[cost.size()+1];
dp[0] = 0;
dp[1] = 0;
//局部，每层最少花费
for(int i=2; i<=cost.size(); i++){
dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2]);
}

return dp[cost.size()];
}
};

python代码

class Solution(object):
def minCostClimbingStairs(self, cost):
"""
:type cost: List[int]
:rtype: int
"""

dp = []
dp.append(0)
dp.append(0)
for i in range(2, len(cost)+1):
dp.append(min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2]))

return dp[len(cost)]